Linked Questions
18 questions linked to/from Dirac equation as Hamiltonian system
4
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0
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242
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Finding Dirac Hamiltonian from Dirac equation [duplicate]
My question is can we get the Hamiltonain from Dirac Equation?
We have the following for Dirac equation:
$$(i\gamma ^\mu \partial_\mu-m)\phi=0.$$
Then separating the time and space components:
$$(...
- 385
1
vote
0
answers
113
views
Why is the anticommutation relation for the Dirac field between fields? [duplicate]
The commutation relation for neutral Klein Gordan field is
$$[\phi(x,t),\pi(x',t)]=i\delta^3(x-x')$$
with all other commutators zero;
The commutation relation for charged Klein Gordan field is
$$[\phi(...
- 859
0
votes
0
answers
65
views
Hamilton's equations for Dirac Hamiltonian [duplicate]
The Dirac Lagrangian
$$\mathcal{L} = i\bar{\psi}\gamma^{\mu}\partial_\mu \psi - m \bar{\psi}\psi$$
gives a Hamiltonian
$$\mathcal{H}(\Pi,\bar{\Pi},\psi,\bar{\psi})=\Pi \dot{\psi}-\mathcal{L}=-\bar{\...
- 781
13
votes
3
answers
4k
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QFT: Propagators are the inverse of the quadratic terms in $\mathcal{L}$?
I am following a QFT course using Peskin & Schroeder (1995): An introduction to Quantum field theory. We've started the functional methods. According to my professor, the vertex rule is the ...
- 1,426
26
votes
1
answer
4k
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Classical Fermion and Grassmann number
In the theory of relativistic wave equations, we derive the Dirac equation and Klein-Gordon equation by using representation theory of Poincare algebra.
For example, in this paper
http://arxiv.org/abs/...
- 1,110
10
votes
1
answer
1k
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How does canonical quantization work with Grassmann variables?
Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.
First, the books introduce the Dirac Lagrangian,
$$\...
- 105k
8
votes
2
answers
2k
views
From Lagrangian to Hamiltonian in Fermionic Model
While going from a given Lagrangian to Hamiltonian for a fermionic field, we use the following formula. $$ H = \Sigma_{i} \pi_i \dot{\phi_i} - L$$ where $\pi_i = \dfrac{\partial L}{\partial \dot{\...
- 1,825
6
votes
1
answer
1k
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Wrong sign anticommutation relation for the Dirac field?
Consider the Dirac Lagrangian $$\mathcal{L}=\psi ^{\dagger }\gamma ^{0}\left(
\mathrm{i}\gamma ^{\rho }\partial _{\rho }-m\right) \psi .$$ The conjugate
momenta to $\psi ^{a}$ are defined, as usual, ...
- 403
3
votes
2
answers
1k
views
Non-relativistic QFT Lagrangian for fermions
Take the ordinary Hamiltonian from non-relativistic quantum mechanics expressed in terms of the fermi fields $\psi(\mathbf{x})$ and $\psi^\dagger(\mathbf{x})$ (as derived, for example, by A. L. Fetter ...
- 31
6
votes
2
answers
409
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Why are Lagrangians linear in $\dot{q}$ so ubiquitous? Gauge theory, Berry phase, Dirac Equation, and more
It seems to me that we encounter first-order equations of motion in some very special situations in physics. It is not clear to me what the connection is, and I am hoping to get some insight into what ...
- 3,780
3
votes
1
answer
1k
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Fermionic Poisson bracket
I'd like to understand the Poisson bracket for fermions in classical field theory defined on a cylinder (with coordinates $(t,x)$, $x$ being the compact direction) and propagating on $T^n$ with ...
- 227
6
votes
1
answer
634
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Non-hermiticity of Dirac Lagrangian: null momentum?
The usual Dirac Lagrangian is $L(\psi,\bar\psi)=\bar\psi(i\not\partial-m)\psi$. The canonical momenta are
$$
\pi=\frac{\partial L}{\partial \psi_{,0}}=i\psi^\dagger \\
\bar \pi=\frac{\partial L}{\...
2
votes
1
answer
622
views
Dirac bracket for the Majorana Lagrangian
Note: See update below.
Consider the Majorana Lagrangian
$$\mathcal{L}=-\psi ^{\mathrm{T}}\mathrm{i}%
\gamma ^{0}\left( \gamma ^{\rho }\partial _{\rho }+m\right) \psi ,\tag{1}$$
where $%
\psi \in ...
- 403
4
votes
2
answers
245
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How to show $\frac{\delta}{\delta \psi(x)}$ being a representation of the operator $\Psi(x)^{\dagger}$ for fermionic Schroedinger Functionals?
I'm following the book of Brian Hatfield, Quantum Field Theory of particles and strings, page 217, eq. 10.89 and the following.
The author is looking for a representation of the operators $\Psi(x)$ ...
- 6,955
1
vote
1
answer
161
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Why does the Hamiltonian of a Lagrangian that consists of only a coupled term become zero?
Let's say our Lagrangian looks something like this:
$$L = \int dz\, Q\cdot \dot{A},\tag{1}$$
where $Q$ and $A$ are two generalized coordinates and $\dot{Q}$ and $\dot{A}$ would be the respective ...
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